If the constant term in the expansion of ${\left(3{\mathrm{x}}^3-2{\mathrm{x}}^2+\frac{5}{{\mathrm{x}}^5}\right)}^{10}$ is $2^{\mathrm{k}}\cdot \mathrm{l}$, where l is an odd integer, then the value of k is equal to :

${\left(3{\mathrm{x}}^3-2{\mathrm{x}}^2+\frac{5}{{\mathrm{x}}^5}\right)}^{10}={\mathrm{x}}^{-50}{\left(3{\mathrm{x}}^8-2{\mathrm{x}}^7+5\right)}^{10}$
constant term in ${\mathrm{x}}^{-50}{\left(3{\mathrm{x}}^8-2{\mathrm{x}}^7+5\right)}^{10}$
coefficient of x⁵⁰ in ${\left(3{\mathrm{x}}^8-2{\mathrm{x}}^7+5\right)}^{10}$
coefficient of x⁵⁰ in $\sum \frac{10!}{{\mathrm{r}}_1!{\mathrm{r}}_{{\mathrm{r}}_1}!{\mathrm{r}}_2!}\left(3{)}^{{\mathrm{r}}_1}\right(-2{)}^{{\mathrm{r}}_2}(5{)}^{{\mathrm{r}}_3}\cdot {\mathrm{x}}^{8{\mathrm{r}}_1+7{\mathrm{r}}_2}$
such that ${\mathrm{r}}_1+{\mathrm{r}}_2+{\mathrm{r}}_3=10$ and $8{\mathrm{r}}_1+7{\mathrm{r}}_2=50$
$\Rightarrow {\mathrm{r}}_1=1;{\mathrm{r}}_2=6\text{ and }{\mathrm{r}}_3=3$
Hence constant term is = $\frac{10!}{1!6!3!}\left(3\right)(-2{)}^6(5{)}^3=\frac{10!}{6!3!}\left(3\right)\left(2{)}^6\right(5{)}^3=2^9\times 5^4\times 21$
Hence k = 9